I don't understand how to graph quadrilateral equations.

## y=16x^2+190x+0

# 2 Answers

**y=16x ^{2}+190x+0**

An algebra II method of graphing this equation is done by the following steps:

**Step 1:****Find the x-coordinate of the vertex and then its correspending y-coordinate**

1. x =^{ -b}/_{2a} is the formula. The a and b apply to your polynomial

2. In your polynomial: a = 16 b = 190 c = 0

3. Hence, x = ^{-(190)}/_{2(16) }----------> x = -5.9375 = ^{-95}/_{16 }

4. Plug this value of x in our polynomial and receive its respective y value. You get the

vertex point.

y=16(^{-95}/_{16})^{2}+190(^{-95}/_{16})+0 ----------> y = -564.0625 = ^{-9025}/_{16}

5. **Vertex ( ^{-95}/_{16},
^{-9025}/_{16})**

**Note:** The** axis of symmetry (which is an imaginary line) runs vertically up and down through this point and this line is defined by x =
^{-95}/_{16}.** This means that the vertex divides the parabola in half. It also means that if you graph a point on one side of the vertex/axis of symmetry, then you will know how this same point will reflect on the other side of the
vertex (keep following the steps to know more)

**Step 2:****Guess two x values that are to the right side of the axis of symmetry/vertex. **

** Find corresponding y values. **

1. The axis is at x = -5.9375. Right of this would be x = 0 and x = 1

2. If x = 0, then y =16(0)^{2}+190(0)+0 = 0 ---->
**(0,0) POINT 1**

If x = 1, then y =16(1)^{2}+190(1)+0 = 206 ---->
**(1,206) POINT 2**

*Step 3:***Reflect POINT 1 and POINT 2 across the axis of symmetry/vertex on the left side
**

** now.**

1. Whenever we reflect points from one side of the vertex to the other side, the y values will remain the same. It is only the x values of the points that will change.

2. The parabola is symmetrical about the axis of symmetry x=
^{-95}/_{16}. If x = 0 (which is 95/16 units to the right of the axis of symmetry), then its reflected x value on the left side of the axis is x =
^{-95}/_{8} (which is ^{95}/_{16} units of the axis of symmetry)
**Reflection of POINT 1 ( ^{-95}/_{8}, 0)**

3. Similarly, if x = 1 (which is 111/16 away from the axis), then its relfected x value on the left side of the axis is x =
^{-103}/_{8 } **Reflection of POINT 2 ( ^{-103}/_{8}, 206)**

_{ }

4. Remember that though the x values were reflected, the y values remained the same.
**Real Life: **Let's say you were waiting at a hotel and there were 2 elevators and you take one of them and your sister takes the other. The metal pole separating the elevators could be
thought of as your y axis and the ground where you waited for the elevators could be your x axis. Now notice that although you both go up to the same height (same y value), your x positions are different. Your elevator's x point could be -1 and your sister's
x position could be 1.

**Step 4:**

symmetric with respect to the imaginary axis of symmetry x =^{-95}/_{16}

Hello Joni!

Set up a table as follows:

X 16*x^2 190*x 0 Y

and then populate the table for a range of x's:

**X 16*x^2 190*x 0 Y**

0 0 0 0 0+0+0=0

1 16 190 0 16+190+0=206

2 64 380 0 64+380+0=444

3 144 570 0 144+570+0=714

etc

You are just calculating the individual powers of each x times its coeficient at the given x in each row of the table. Include a sufficient number of x's to disclose the behavior of the function. This function has an x^2 so it is a parabola with a minimum value at some x. When you are satisfied you have enough x's calculated (and find the minimum Y result) you just plot the pairs of x and y values, ie the first and last column values.

A graphing calculator does just this type of calculation. I don't know if you are using a graphing calculator but if you do it manually as described above you will get better appreciation for the function's characteristics and an even stronger appreciation of how slick a graphing calculator can be.

Good Luck! *BruceS*